Answer
$$\int^1_0\frac{dx}{(x+1)(x^2+1)}=\frac{1}{4}\ln2+\frac{\pi}{8}$$
Work Step by Step
$$A=\int^1_0\frac{dx}{(x+1)(x^2+1)}$$
1) Express the integrand as a sum of partial fractions:
Since $x^2+1$ is an irreducible quadratic factor,
$$\frac{1}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$$
Clear fractions:
$$Ax^2+A+Bx^2+(B+C)x+C=1$$ $$(A+B)x^2+(B+C)x+(A+C)=1$$
Equating coefficients of corresponding powers of $x$, we get
- $A+B=0$
- $B+C=0$
- $A+C=1$
Solving for $A$, $B$, $C$, we get $A=1/2$, $B=-1/2$ and $C=1/2$.
Therefore,
$$\frac{1}{(x+1)(x^2+1)}=\frac{1}{2}\frac{1}{x+1}-\frac{1}{2}\frac{x-1}{x^2+1}$$
2) Evaluate the integral:
$$A=\frac{1}{2}\int^1_0\frac{dx}{x+1}-\frac{1}{2}\int^1_0\frac{x-1}{x^2+1}dx$$ $$A=\frac{1}{2}\ln|x+1|\Big]^1_0-\frac{1}{2}\Big(\int^1_0\frac{x}{x^2+1}dx-\int^1_0\frac{1}{x^2+1}dx\Big)$$ $$A=\frac{1}{2}\ln2-\frac{1}{2}\Big(\frac{1}{2}\int^1_0\frac{1}{x^2+1}d(x^2+1)-\tan^{-1}x\Big]^1_0\Big)$$ $$A=\frac{1}{2}\ln2-\frac{1}{2}\Big(\frac{1}{2}\ln|x^2+1|\Big]^1_0-\frac{\pi}{4}\Big)$$ $$A=\frac{1}{2}\ln2-\frac{1}{4}\ln2+\frac{\pi}{8}$$ $$A=\frac{1}{4}\ln2+\frac{\pi}{8}$$
